1984 AJHSME Problem 9

Revision as of 19:41, 31 January 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem == The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math> <math>\text{(A)}\ \...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

The product of the 9 factors $\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =$

$\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}$

Solution

This product simplifies to: \[\frac{1}{2} \cdot \frac{2}{3} \dots \frac{9}{10}.\] Numerators and denominators cancel to yield the answer: $\boxed{\text{(A)} \frac{1}{10}}.$